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Lorenz formulas

  1. Jun 7, 2005 #1
    In the time lorenz formulas, how can u tell what the difference between t-zero and T is?
    t-zero
    T= --------------------
    -------------------
    | 1 - (v^2/c^2)
    \|

    for example in the following problem:
    With what speed will a clock have to be moving in order to run at rate that is one-half the rate of a clock at rest?

    How can u know what to plug in for t-zero and T?

    Thanks so much
     
  2. jcsd
  3. Jun 7, 2005 #2
    I have always found it very helpful in these cases for follow the derivation of the formula. Then you'll understand it much better. See - http://www.geocities.com/physics_world/sr/light_clock.htm

    Pete
     
  4. Jun 7, 2005 #3
    Hint: your T and t-zero are not points in time, but time intervals.
     
  5. Jun 7, 2005 #4
    thank u for the tips yet still I get the wrong number when I plug my numbers in:

    -->With what speed will a clock have to be moving in order to run at rate that is one-half the rate of a clock at rest?

    For the problem, I plug in T=(1/2)t and t-zero=t, because I think the moving clock the time is one half-t, while to the clock at rest the time is t. Doesn't t-0 represent what the clock at rest observes??

    thanks for any further help anyone provides
     
  6. Jun 7, 2005 #5

    Doc Al

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    Staff: Mentor

    [itex]T_0[/itex] represents the time as measured by the "moving" clock; [itex]T[/itex] represents the time as measured by "stationary" clocks.

    [tex]T = \frac{T_0}{\sqrt{1 - v^2/c^2}}[/tex]

    In your example, if the moving clock measures t seconds, then the stationary observers will measure 2t seconds according to their clocks. (They measure the moving clock to be running at half the normal rate.)
     
  7. Jun 7, 2005 #6

    robphy

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    This mysterious and often confusing formula of time-dliation can interpreted geometrically.
    As Dodo says, "Hint: your T and t-zero are not points in time, but time intervals."


    Let's introduce events:
    the common meeting event O,
    the distant event t0 on the moving clock [when the moving clock reads t0],
    the local event T [when the stationary clock reads T], which our stationary observer says is simultaneous with the distant event t0. (Note that the moving observer does not regard these two events as simultaneous!)

    OT and Ot0 are [timelike] legs of a [Minkowski]-right triangle on a spacetime diagram (as drawn below.... time runs upwards by convention).

    The legs of the triangle are OT, Ot0 and Tt0.
    Ot0 is the "hypotenuse" of this [Minkowski]-right triangle.
    OT is the "adjacent side" and Tt0 is the "opposite side"... these legs are [Minkowski]-perpendicular.
    Another way to describe this is that the vector Ot0 is being resolved into temporal and spatial components by our stationary observer.
    [By the way, the "angle" [tex]\theta [/tex] is called the rapidity... and it is related to the relative velocity by [tex]v=c\tanh\theta[/tex]. ]

    So, think this way:

    (adjacent side OT) = "COSINE(ANGLE)" (hypotenuse Ot0),
    where "COSINE(ANGLE)" in the Minkowski geometry is [tex]\cosh\theta=\frac{1}{\sqrt{1-\tanh^2\theta}}=\frac{1}{\sqrt{1-(v/c)^2}}=\gamma[/tex]

    [tex]\begin{picture}(200,200)(0,0)
    \put(50,30){O}
    \put(50,50){\textcolor{red}{\line(3,4){52}}}
    \put(50,50){\textcolor{green}{\line(0,1){70}}}
    \put(50,120){\textcolor{green}{\line(1,0){50}}}
    \put(50,125){T}
    \put(100,125){\[t_0\]}
    \put(48,65){\[\theta\]}
    \end{picture}
    [/tex]

    A spacetime diagram is worth a thousand words.
     
  8. Jun 7, 2005 #7
    That is quite true Rob. Someone asked a question here a while back which was best answered with a spacetime diagram. Here is the answer I gave to the question given at the top of the page

    http://www.geocities.com/physics_world/sr/st_diagram.htm

    I think the questioner was confusing spatial distances with spacetime intervals.

    Pete
     
  9. Jun 7, 2005 #8

    robphy

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    It's a mystery to me why "spacetime diagrams" are rarely found in the relativity section of introductory textbooks.... but they're everywhere in the [Galilean] kinematics section! They really can clear up a lot of ambiguities and misunderstandings.
     
  10. Jun 7, 2005 #9
    Good question. Let me see what I can learn on that. It would seem like a natural step since newer texts are introducing the spacetime interval and the spacetime diagram is awesome for describing that.

    Pete
     
    Last edited: Jun 7, 2005
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